Question

Two balanced three-phase loads, as shown in the figure, are connected to a 100$\sqrt{3}$ V three-phase, 50 Hz main supply. Given \( Z_1 = (18 + j24) \, \Omega \) and \( Z_2 = (6 + j8) \, \Omega \), the ammeter reading, in amperes, is ________. (round off to nearest integer)

Hint:

For a balanced three-phase system, the total current can be calculated by multiplying the phase current by \( \sqrt{3} \).

Answer 1

Correct Answer: 20

In a balanced three-phase system, the current in each phase can be calculated using the formula: \[ I = \frac{V_{\text{phase}}}{Z_{\text{total}}} \] Where: - \( V_{\text{phase}} = \frac{V_{\text{line}}}{\sqrt{3}} \) is the phase voltage,
- \( Z_{\text{total}} \) is the total impedance per phase.
Given: - \( V_{\text{line}} = 100 \sqrt{3} \, \text{V} \), so \( V_{\text{phase}} = 100 \, \text{V} \),
- \( Z_1 = (18 + j24) \, \Omega \) and \( Z_2 = (6 + j8) \, \Omega \),
- The total impedance per phase is the sum of the impedances: \[ Z_{\text{total}} = Z_1 + Z_2 = (18 + j24) + (6 + j8) = (24 + j32) \, \Omega.
\] The magnitude of \( Z_{\text{total}} \) is: \[ |Z_{\text{total}}| = \sqrt{24^2 + 32^2} = \sqrt{576 + 1024} = \sqrt{1600} = 40 \, \Omega.
\] Now, calculate the current per phase: \[ I_{\text{phase}} = \frac{100}{40} = 2.5 \, \text{A}.
\] Since the system is three-phase, the total current \( I_{\text{total}} \) is: \[ I_{\text{total}} = I_{\text{phase}} \times \sqrt{3} = 2.5 \times \sqrt{3} \approx 4.33 \, \text{A}.
\] Thus, the ammeter reading is approximately \( \boxed{4} \) A.